Are the convergence radii circles of a Laurent-series always caused by isolated singularities? Laurent series
$$f(z) := \sum_{n=-\infty}^\infty a_n (z-c)^n$$
converge for $r<|z-c|<R$ where
$$r = \limsup_{n\to\infty}|a_{-n}|^{\frac1n},
\\\frac1R = \limsup_{n\to\infty}|a_n|^{\frac1n}.$$
But what lies on the boundary of such an annulus that breaks convergence? If $r\neq R$, is it always isolated singularities (of the analytical continuation; essential or not?), or could something else also be the cause?
Or asked differently, is such an $f(z)$ always analytically continuable for almost all $|z|\in\{r,R\}$?
 A: For an open set $U \subset \mathbb{C}$, there always are $f \in \mathcal{O}(U)$ which cannot be continued analytically across any part of the boundary $\partial U$.
That is an easy consequence of the (general) Weierstraß product theorem, which I quote here from Rudin (Real and Complex Analysis; Thm 15.11):

Let $\Omega$ be an open set in $S^2$, $\Omega \neq S^2$. Suppose $A \subset \Omega$ and $A$ has no limit point in $\Omega$. With each $\alpha \in A$ associate a positive integer $m(\alpha)$. Then there exists an $f \in H(\Omega)$ all of whose zeros are in $A$, and such that $f$ has a zero of order $m(\alpha)$ at each $\alpha \in A$.

With that, all one needs is the existence of a set $A$ contained in the annulus $K = \{ z : r < \lvert z\rvert < R\}$ such that $A$ has no limit point in $K$, and every boundary point of $K$ is a limit point of $A$. Then, choosing all multiplicities as $1$ for simplicity, one obtains a function $f\not\equiv 0$ such that every boundary point is a limit of zeros of $f$, hence $f$ cannot be analytically continued across any part of the boundary with isolated singularities, since such a continuation would vanish on a non-discrete set without vanishing identically, contradicting the identity theorem.
For the special case of annuli, we can construct such a function relatively explicitly.
For $n \in \mathbb{Z}^+$, let $\alpha_n = \left(1 - \frac{1}{(n+1)^2}\right)e^{in}$. Then $(\alpha_n)_{n \in \mathbb{Z}^+}$ is a sequence in the unit disk $\mathbb{D}$ whose limit points are exactly the points on the unit circle $\partial \mathbb{D}$. (That no other points are limit points is immediate, that every point on the unit circle is a limit point follows from the fact that $A_k := \{e^{in} : n \geqslant k \}$ is dense in the unit circle for all $k\in \mathbb{Z}^+$.)
Then consider the Blaschke product
$$B(z) := \prod_{n=1}^\infty \frac{\alpha_n - z}{1 - \overline{\alpha_n}z}\frac{\lvert \alpha_n\rvert}{\alpha_n}.$$
Since $\sum (1 - \lvert\alpha_n\rvert) = \pi^2/6 - 1 < \infty$, $B \in H^\infty(\mathbb{D})$ and $B$ has simple zeros in each $\alpha_n$ and no other zeros, by the general theory of Blaschke products.
Now, if $0 < r < R < \infty$, the function
$$f(z) = B\left(\frac{z}{R}\right)\cdot B\left(\frac{r}{z}\right)$$
is holomorphic on the annulus and has simple zeros [$R\cdot\alpha_k = r/\alpha_m$ can't happen, because $e^{i(k+m)} \neq 1$ for $k,\,m \in \mathbb{Z}^+$, so the zeros remain simple] only in the points $R\cdot\alpha_n$ and $r/\alpha_n$ for $n$ sufficiently large, so that $R\lvert\alpha_n\rvert > r$ resp. $r/\lvert\alpha_n\rvert < R$. Every point on $\lvert z\rvert = R$ is a limit point of the $R\cdot\alpha_n$, and every point on $\lvert z\rvert = r$ is a limit point of the $r/\alpha_n$, so every boundary point of the annulus is a limit point of zeros of $f$.
If $r = 0$, then $0$ is trivially an isolated singularity of each $g \in \mathcal{O}(K)$, and if $R = \infty$, then $\infty$ is trivially an isolated singularity of each $g \in \mathcal{O}(K)$. If $r = 0$ and $R = \infty$, then each boundary point of $K$ is an isolated singularity of all $g \in \mathcal{O}(K)$, if only one of these equations hold then drop the corresponding $B(z/R)$ or $B(r/z)$ from $f$ to obtain a function that cannot be analytically continued (with isolated singularities) across any part of the boundary.
A: Yes.
Hint: Otherwise you could extend the function to larger domain using the compactness of the circle.
